I coulnd't find clear explanations about the production of Iodine-131 in nuclear reactors based on U-235.

  • Can it be derived directly from U-235 ? In this case : U-235 + neutron -> I-131 + what ???

  • Or is it derived indirectly from U-235 ? In this case : U-235 + neutron -> X ? -> I-131 (what could be X? And by which process? neutron impact on X or beta-decay of X?)

Thanks for any information.


It is a fission product, one of many nuclides: http://hyperphysics.phy-astr.gsu.edu/hbase/NucEne/fisfrag.html

As uranium has Z=92 and iodine Z=53, the other piece could be yttrium (Z=39). But total Z may change because of all the beta-decays. The iodine isotope is likely a decay product of short-lived fission products like In-131, Sn-131, Sb-131, Te-131.

  • $\begingroup$ Hi Pieter, in your link, there is a table "Table of fission fragments" in which I-131 is not listed as a possible direct fission product of U-235. So, I-131 does appear, of course, but the chain of reactions leading to it remains a mystery... $\endgroup$ – Andrew Dec 20 '16 at 15:59
  • $\begingroup$ Uranium, like most heavy elements has a higher N/Z than lighter elements, at least their stable isotopes. Of course, neutron-induced fission of U-235 produces 2 extra neutrons, but you still have quite an excess. Z will increase, in the decay chains, to bring N/Z down to stable levels via beta decay. I'd look at e.g. Sn-131 (N=50) and Mo-102 (N=42), both unstable under beta decay. (Note that N adds up to 92, Z to 233. U-235 of course has Z=235 but fission produces neurons). Sn-131 will decay to I-131. $\endgroup$ – MSalters Dec 20 '16 at 16:04
  • $\begingroup$ The thesis you will find here : « esirc.emporia.edu/handle/123456789/3523 » looks quite interesting. Look at page 89 ; it gives a way for directly producing I-131 : U-235 + 1n -> U-236 -> Y-101 + I-131 + 4n Surely, many other ways are possible, through beta-decay of Te, Sn or else ; but at least, I have something to eat… $\endgroup$ – Andrew Dec 20 '16 at 17:54

Uranium only serves as a neutron source. Yes, the decay reaction produces I-131, but it's in the radioactive waste, and it would be a pain to extract it. Your typical reactor fuel is left in for many years, during which quite a bit of the produced I-131 would decay again. Plus, the waste is quite radioactive, making it inconvenient to handle. And you get a lot of I-129 that you can't chemically separate.

Actual production is based on a convenient fact: Natural Tellurium can be radiated with neutrons from U-235 fission. This will produce a wide variation of isotopes, but if you give the short-lived isotopes time to decay (~few hours), let the Xenon escape, and chemically remove the remaining Tellurium, you'll find that the remaining mixture is very rich in I-131.

  • $\begingroup$ Ah, now I understand the question, this was about producing isotopes for medical purposes etc. $\endgroup$ – user137289 Dec 19 '16 at 20:22
  • $\begingroup$ Could you write a reaction (or a chain of reactions) starting with U-235 and leading to I-131? I'm sorry to say, but it seems that nobody knows... $\endgroup$ – Andrew Dec 19 '16 at 20:29
  • $\begingroup$ @Andrew: The point of this answer is, that's not how you do it. Uranium is there to produce neutrons, the bulk of the Z for iodine-131 comes from Te-130. (Via Te-131) $\endgroup$ – MSalters Dec 19 '16 at 21:37
  • $\begingroup$ OK. I found : Te-130 + neutron -> Te-131 -> I-131 by beta decay. Now, do we have U-235 -> ... -> Te-130 ? $\endgroup$ – Andrew Dec 20 '16 at 8:39
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    $\begingroup$ @Andrew: Right, that seems to be a major misunderstanding there. I-131 is intentionally produced, e.g. for medical reasons, in the way that I describe (Tellurium on the side). But when you refer to Chernobyl & Fukushima, you're talking about the unintentional radioactive waste. See Pieter's answer for that. $\endgroup$ – MSalters Dec 20 '16 at 15:38

Most radiopharmaceuticals are accelerator produced. Less costly than building a reactor. Accelerators offer a lower radiation risk to workers and essentially zero risk to the public.

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    $\begingroup$ I'm not sure how this answers the question though. Can you please elaborate? $\endgroup$ – AccidentalFourierTransform Jun 20 '18 at 22:11

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